Given a binary search tree and the lowest and highest boundaries as L
and R
, trim the tree so that all its elements lies in [L, R]
(R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.
Example 1:
Input:
1
/ \
0 2
L = 1
R = 2
Output:
1
\
2
Example 2:
Input:
3
/ \
0 4
\
2
/
1
L = 1
R = 3
Output:
3
/
2
/
1
解法1:
凡是树的题目无非DFS,BFS,递归或者迭代做,仅此而已。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def trimBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: TreeNode
"""
# use DFS, root first then left and right
# if L <= root.val <= R:
# new_root with root.val;
# new_root.left = self.trimBST(root.left, L, root.val)
# new_root.right = self.trimBST(root.right, root.val, R)
# return new_root
# elif root.val < L:
# return self.trimBST(root.right, L, R)
# elif root.val > R:
# return self.trimBST(root.left, L, R)
if root is None:
return None
if L <= root.val <= R:
new_root = TreeNode(root.val)
new_root.left = self.trimBST(root.left, L, root.val)
new_root.right = self.trimBST(root.right, root.val, R)
return new_root
elif root.val < L:
return self.trimBST(root.right, L, R)
elif root.val > R:
return self.trimBST(root.left, L, R)
迭代解法:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if (root == null) {
return root;
}
while (root.val < L || root.val > R) {
if (root.val < L) {
root = root.right;
}
if (root.val > R) {
root = root.left;
}
}
TreeNode dummy = root;
while (dummy != null) {
while (dummy.left != null && dummy.left.val < L) {
dummy.left = dummy.left.right;
}
dummy = dummy.left;
}
dummy = root;
while (dummy != null) {
while (dummy.right != null && dummy.right.val > R) {
dummy.right = dummy.right.left;
}
dummy = dummy.right;
}
return root;
}
}
标签:Trim,Binary,Search,right,TreeNode,val,dummy,root,left From: https://blog.51cto.com/u_11908275/6381143